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it is given that dy = 3y. You can take it from there.
After you get that, do you integrate both sides?
the problem is not clear. does y change at a rate exactly three times its previous value?
I would assume so because it does say values...does that help any?
well, if y(1) = 4e, then y(2) = 3 times 4e and y(3) = 3 times 3times 4e and so on
hmm so that would be 105e? That doesn't feel right, but I mean the steps make sense.
otherwise it would be dy = 3y ==> y = 3y2 +c 3y^2 -y +c = 0. substitute y(1) = 4e and solve for c then use that to find y(5)
no wait. that is wrong. scrap that.
The second one or both?
the second one.
So, the first one was indeed right?
I think so. Is this problem in application of derivatives or integrals or something?
they have given y(1) = 4e. what is the 1? is it y at 1 second? is y a function of time?
I'm not really sure...I'm just doing a huge packet of practice problems to help me study for the AP calculus test.
I can't see it being a function of time...There is not information relating the problem to time.
where rate is given, it is generally a function of time.
Oh, I completely overlooked the rate of it all.
so dy(t)/dt = 3y(t-1) ?
Yes, so is that just the original function of y or is this y at 1?
no, it is the general function of y.
Okay, but why is the variable "y" in the function? It don't think that y should be in there. It's like putting a f in the function.
because it is given that it changes at 3 times its values.
yes, a function can depend on its previous values. you can put an f in the function.
hmmm...well how would be begin to figure out y(5)'s function given that y(1)=4e? There is no "t" in "4e"..
no no, t = 1.
I think my first method was correct. there is no other explanation with the data given. so yeah. I guess 105e is your answer.
just post your question and your answer, 105 e in a new post and see what everyone else thinks of the answer.
Okay, that sounds better. I felt like through the second route, we were reading too much into the question...hmmm maybe...idk...I will try it. I do appreciate all of your help though, Thank you so much.
you are welcome.